Which of the following is true for the definite integrals shown above? $$J = \int_{0}^{1}\sqrt{1-x^4}\,dx$$
$$K = \int_{0}^{1}\sqrt{1+x^4}\,dx$$
$$L = \int_{0}^{1}\sqrt{1-x^8}\,dx$$
Which of the following is true for the definite integrals shown above?
(A) $J<L<1<K$
(B) $J<L<K<1$
(C) $L<J<1<K$
(D) $L<J<K<1$
(E) $L<1<J<K$
What is the smartest way of solving this question other than solving the integrals as this question must take from me at most 2.5 minutes during the examination?  
 A: Since $\sqrt{x}$ is increasing, any inequality among the radicands holds also for the integrals. On $[0,1]$ we have $0 \leq x^8 \leq x^4 \leq 1.$ So $$1-x^4 \leq 1-x^8 \leq 1 \leq 1+x^4.$$
Edit: And obviously none of these is equality everywhere on the interval! Thus the strict inequalities in the given statement.
A: $\begin{align}L-J&=\int_0^1 \sqrt{1-x^8}dx-\int_0^1 \sqrt{1-x^4}dx\\
&=\int_0^1 \left(\sqrt{1-x^8}-\sqrt{1-x^4}\right)dx\\
&=\int_0^1 \dfrac{\left(\sqrt{1-x^8}\right)^2-\left(\sqrt{1-x^4}\right)^2}{\sqrt{1-x^8}+\sqrt{1-x^4}}dx\\
&=\int_0^1 \dfrac{x^4(1-x^4)}{\sqrt{1-x^8}+\sqrt{1-x^4}}dx>0
\end{align}$
Therefore $L>J$ and $(C),(D),(E)$ are false.
$\begin{align} K-1&=\int_0^1 \sqrt{1+x^4}dx-\int_0^1 1 dx\\
&=\int_0^1 \left(\sqrt{1+x^4}-1\right)dx\\
&=\int_0^1 \dfrac{\left(\sqrt{1+x^4}\right)^2-1^2}{\sqrt{1+x^4}+1}dx\\
&=\int_0^1 \dfrac{x^4}{\sqrt{1+x^4}+1}dx>0\\
\end{align}$
Therefore $K>1$ and $(B)$ is false.
Therefore $(A)$ is true
I have supposed the question has an unique answer.
Otherwise to be sure $(A)$ is the good answer it remains to prove that $L<1$. Same method can be used to prove this.
